Question: You have found the following ages (in years) of all 5 porcupines at your local zoo: $ 6,\enspace 10,\enspace 19,\enspace 6,\enspace 12$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{6 + 10 + 19 + 6 + 12}{{5}} = {10.6\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $6$ years $-4.6$ years $21.16$ years $^2$ $10$ years $-0.6$ years $0.36$ years $^2$ $19$ years $8.4$ years $70.56$ years $^2$ $6$ years $-4.6$ years $21.16$ years $^2$ $12$ years $1.4$ years $1.96$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{21.16} + {0.36} + {70.56} + {21.16} + {1.96}} {{5}} $ $ {\sigma^2} = \dfrac{{115.2}}{{5}} = {23.04\text{ years}^2} $ The average porcupine at the zoo is 10.6 years old. The population variance is 23.04 years $^2$.